How much current is flowing in a wire 4.55 m long if the maximum force on it is 0.575 N when placed in a uniform 0.0700-T field?

1.81A

The magnitude of the magnetic force (F) acting on a current carrying wire is given by;

F = I x L x B sin θ      (i)

Where;

I = current on the wire

B = magnetic strength

L = length of the wire

θ = angle between the wire and the magnetic field

From the question, there is a maximum force (F = 0.575N) of on the wire. This means that the angle between the wire and the magnetic field is 90°.

Also;

the magnetic field(B) = 0.0700-T

the length(L) of the wire = 4.55m

Substitute these values into equation (i) as follows;

0.575 = I x 4.55 x 0.0700 x sin 90°

0.575 = I x 4.55 x 0.0700 x 1

0.575 = I x 0.3185

Solve for I;

I = 0.575 / 0.3185

I = 1.81A

Therefore, the current flowing in the wire is 1.81A

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr